Improper Integrals and Their Convergence
Give meaning to an integral with an infinite limit or an unbounded integrand by evaluating a proper integral and taking a limit, then decide whether the result is finite.
What a learner can do afterwards
- Rewrite infinite-range integrals with separate limits where needed, evaluate them and find a cutoff from a tail error
- Use the convergence threshold and finite-value formula for 1/x^p on [1, infinity), including finding p from a given value
- Handle endpoint and interior singularities with the required separate one-sided limits
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Use your antiderivatives, including partial fractions, with finite cutoffs. An improper integral has an infinite bound or an integrand unbounded near an endpoint or interior point. Convergence requires every cutoff limit to be finite. For f continuous on [A, infinity), use lim(t -> infinity) ∫A^t f(x) dx. For f continuous on (-infinity,B], use lim(a -> -infinity) ∫a^B f(x) dx. Evaluate each proper integral before taking its limit.
For ∫2^infinity 4/x^2 dx, an antiderivative is -4/x. Integrating from 2 to t gives 2-4/t, whose limit is 2. For comparison, ∫2^t 1/x dx = ln(t)-ln(2), which grows without bound. The first improper integral converges; the second diverges. An integrand tending to 0 does not by itself establish convergence.
For real p, ∫1^infinity x^(-p) dx converges exactly when p > 1. For p != 1, the integral to t is (t^(1-p)-1)/(1-p). If p > 1, t^(1-p) tends to 0, leaving 1/(p-1). If p < 1, the expression grows without bound. At p = 1, it is ln(t), which also grows without bound. To find p from a finite value V, solve 1/(p-1)=V.
For c > 0 and t > 0, the omitted tail ∫t^infinity c/x^2 dx equals c/t. Using -c/x as an antiderivative, the integral from t to b is c/t-c/b; let b grow to obtain c/t. When approximating ∫A^infinity by ∫A^t, with 0 < A < t, this tail is the error. For an exact error E > 0, solve c/t=E to get t=c/E.
Use one-sided cutoffs at singular endpoints. For ∫0^B 1/sqrt(x) dx with B > 0, integrate from a > 0: 2sqrt(B)-2sqrt(a). Let a -> 0 from above to get 2sqrt(B). At the right endpoint of ∫0^L 1/sqrt(L-x) dx, take b < L. The antiderivative -2sqrt(L-x) gives 2sqrt(L)-2sqrt(L-b). Let b -> L from below to get 2sqrt(L), for L > 0.
If f is unbounded near c inside [A,B] and continuous elsewhere, split at c. Evaluate lim(u -> c from below) ∫A^u f(x) dx and lim(v -> c from above) ∫v^B f(x) dx independently. Add their values only if both are finite. Opposite infinities cannot cancel to define a convergent improper integral.
For 1/sqrt(|x-c|), the left antiderivative is -2sqrt(c-x), and the right is 2sqrt(x-c). On [A,B], with A < c < B, the separate limits are 2sqrt(c-A) and 2sqrt(B-c). Both are finite. For example, on [-9,16] with c=0, the limits are 6 and 8, giving 14.
For f continuous on the real line, split ∫-infinity^infinity f(x) dx at 0. Take the left limit a -> -infinity and the right limit b -> infinity independently. Add only if both integrals converge. Symmetric cutoffs alone do not decide convergence: ∫-t^t x dx = 0, but the left integral is -t^2/2 and the right is t^2/2. Each is unbounded as t grows, so the improper integral diverges.
Use finite cutoffs at infinite bounds and singularities. Evaluate each required limit separately. Add the values only when every required limit is finite.
Lesson and questions written for this dataset. Calculus texts and problem records were consulted for mathematical facts and question design.
Readings and practice questions written for this dataset. Linked video: Khan Academy, Introduction to improper integrals. The video's license has not been verified.
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Which endpoint moves?
The direction of a cutoff depends on the interval. For an infinite upper bound, keep the lower endpoint fixed and increase a finite upper endpoint. For a negative infinite lower bound, keep the upper endpoint fixed and move a finite lower endpoint toward negative infinity.
For example, integrate 3/(x+2)^2 from 0 to a finite t. The antiderivative -3/(x+2) gives 1.5-3/(t+2). The subtracted term tends to zero as t grows, so the improper integral from 0 to infinity is 1.5.
For the other direction, consider 2/(1-x)^2 from negative infinity to 0. On this interval an antiderivative is 2/(1-x). Starting at a finite a<0 gives 2-2/(1-a). As a tends to negative infinity, this expression tends to 2.
Both calculations use ordinary endpoint subtraction on finite intervals first. The final limit determines whether the improper integral has a finite value. If there are two infinite bounds, calculate the two tails independently.
A singularity needs its own limit
A function becoming unbounded near an endpoint does not decide whether its integral converges. For 1/sqrt(11-x) from 2 to 11, stop at b<11. Using -2sqrt(11-x) as an antiderivative gives 6-2sqrt(11-b). This tends to 6 as b approaches 11 from below. The integral is finite although the function is unbounded near 11.
An interior singularity requires a separate calculation on each side. Consider 1/(x-3) on [1,6]. On the left of 3, the integral from 1 to u is ln(3-u)-ln(2). It tends to negative infinity as u approaches 3 from below. On the right, the integral from v to 6 is ln(3)-ln(v-3). It tends to positive infinity as v approaches 3 from above.
Neither limit is finite, so this improper integral diverges. Choosing related cutoffs to cancel the two unbounded terms would test a different limit. It would not establish convergence of either required piece.
Values, signs and omitted tails
A definite integral gives signed accumulation. For instance, integrating the constant function -2 from 0 to 3 gives -6, while the geometric area between that graph and the axis is 6. For a continuous function on a finite closed interval, an antiderivative F gives the integral by F(B)-F(A). Some bounded functions with jumps are also integrable; continuity is sufficient, not necessary.
For an improper integral, perform the finite-interval calculation before taking each limit. The family x^(-p) on [1,infinity) has finite value 1/(p-1) when p>1. At p=1 the finite integral is ln(t), which grows without bound. Values p<1 also give divergence.
A convergent integral can still differ from a finite approximation. For c>0, omitting everything beyond a positive cutoff t from an integral of c/x^2 leaves a tail of c/t. If the desired omitted amount is E>0, the equation c/t=E gives t=c/E. Check that this cutoff lies beyond the starting point. The error here is the omitted integral, not the height of the function at t.
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